What is area x, 1/(x^2),2 ?

The answer to area x, 1/(x^2),2 is diverges-2sqrt(2)+7/2

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area x, 1/(x^2),2

area

x, \frac{1}{x ^{2}}, 2

diverges - 22 + \frac{7}{2}

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

x = \frac{1}{x ^{2}} : x = 1

x = 2 : x = 2

\frac{1}{x ^{2}} = 2 : x = \frac{1}{2}, x = - \frac{1}{2}

= \int_{- \frac{1}{2}}^{\frac{1}{2}} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{\frac{1}{2}}^{1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{1}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{x ^{2}} - 2 d x + \int_{\frac{1}{2}}^{1} \frac{1}{x ^{2}} - 2 d x + \int_{1}^{2} ∣ x - 2 ∣ d x

\int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{x ^{2}} - 2 d x =

diverges

\int_{\frac{1}{2}}^{1} \frac{1}{x ^{2}} - 2 d x = 3 - 22

\int_{1}^{2} ∣ x - 2 ∣ d x = \frac{1}{2}

= diverges + 3 - 22 + \frac{1}{2}

Simplify

= diverges - 22 + \frac{7}{2}

\frac{d}{d a} (sin^{4} (a) - cos^{4} (a))

\frac{1}{4} y^{^{''}} + 4 y = 2 tan (4 t) - \frac{1}{3} e^{3 t}

\int (1 + y^{2}) d y

\frac{\partial}{\partial x} (3 x \cdot e^{z})

What is area x, 1/(x^2),2 ?
The answer to area x, 1/(x^2),2 is diverges-2sqrt(2)+7/2